* Let $a(n)$ be [A069999][1] (i.e., number of possible dimensions for commutators of $n \times n$ matrices; it is independent of the field). OEIS states that no generating function is known.
* Let $P(n,k)$ be a triangle read by rows such that
$$
P(n, k) = \begin{cases}
0 & \textrm{if } k > n \\
1 & \textrm{if } k = 0 \\
z^{n-k+1}P(n, k-1) + P(k-1, k-1) & \textrm{otherwise}
\end{cases}
$$

I conjecture that number of positive terms in $P(n-1, n-1)$ equals $a(n)$.

Here is the PARI/GP program to generate number of positive terms of $P(n-1, n-1)$:

    upto1(n) = my(v1); v1 = vector(n, i, 1); for(i=1, n-1, for(j=i+1, n, v1[j] = v1[i] + z^(j-i)*v1[j])); v1 = vector(n, i, #select(x->x>0, Vec(v1[i])))

Is there a way to prove it? Is it possible to get a closed-form or generating function after some manipulations with $P(n-1,n-1)$?

  [1]: https://oeis.org/A069999